Question

Solve $\sin$ $\left(t\right)=\cos \left(2t\right)$ for $-4\pi \le t\le -2\pi$

• A. $\left\{\begin{array}{c}\frac{-\pi }{6},\frac{5\pi }{6},\frac{-\pi }{2}\end{array}\right\}$
• B. $\left\{\begin{array}{c}\frac{-\pi }{3},\frac{-5\pi }{3},\frac{-\pi }{2}\end{array}\right\}$
• C. $\left\{\begin{array}{c}\frac{-23\pi }{6},\frac{-19\pi }{6},\frac{-5\pi }{2}\end{array}\right\}$
• D. $\left\{\begin{array}{c}\frac{-23\pi }{3},\frac{-19\pi }{3},\frac{-5\pi }{2}\end{array}\right\}$
• E. $\left\{\begin{array}{c}\frac{-11\pi }{3},\frac{-7\pi }{3},\frac{-5\pi }{2}\end{array}\right\}$

Answer 1

The correct option is C $\left\{\begin{array}{c}\frac{-23\pi }{6},\frac{-19\pi }{6},\frac{-5\pi }{2}\end{array}\right\}$

We know that $\cos \left(2t\right)=1-2\sin {}^2\left(t\right)$

Substitute $\cos \left(2t\right)=1-2\sin {}^2\left(t\right)$ in the equation $\sin \left(t\right)=\cos \left(2t\right)$ and find the roots:

$\sin \left(t\right)=1-2\sin {}^2\left(t\right)\Rightarrow 2\sin {}^2\left(t\right)+\sin \left(t\right)-1=0\Rightarrow 2\sin {}^2\left(t\right)+2\sin \left(t\right)-\sin \left(t\right)-1=0\Rightarrow 2\sin \left(t\right)(\sin \left(t\right)+1)-1(\sin \left(t\right)+1)=0\Rightarrow 2\sin \left(t\right)-1=0,\sin \left(t\right)+1=0\Rightarrow \sin \left(t\right)=\frac{1}{2},\sin \left(t\right)=-1$

In the interval $0\le t\le 2\pi$, the answer to this problem would be $\{\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}\}$.

Subtract $4\pi$ from $\{\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}\}$ to satisfy the domain restriction of the problem that is

$\{\frac{-23\pi }{6},\frac{-19\pi }{6},\frac{-5\pi }{2}\}$